Saturday

December 20, 2014

December 20, 2014

Posted by **anon** on Saturday, March 12, 2011 at 3:06am.

A.3e^-2x sec(3x) tan (3x)-2e^-2x Sec(3x)<<<< my choice

B.)3e^-2x sec(3x)tan (3x)-2xe^-2x sec(3x)

C.)3e^-2x sec(3x)tan (x)-2e^-2x Sec(3x)

D.)3e^-2x sec(3x)tan (x)-2xe^-2x sec(3x)

2.)calculate dy/dx if y=Ln(2x^3+3x)

A.)1/2x^3+3x

B.)1/6x^2+3

C.)2x^3+3x/6x^2+3

D.)6x^2+3/2x^3+3x <<<< my choice

3.)Given the curve that is described by the equation r=3 cos theta, find the angle that tangent lines makes with the radius vector when theta=120.

A.) 30deg

B.) 45deg

C.) 60deg

D.) 90deg << my choice

4.)A line rotates in a horizontal plane according to the equation theta=2^t-6t,where theta is the angular position of the rotating line, in radians ,and tis the time,in seconds. Determine the angular acceleration when t=2sec.

A.) 6 radians per sec^2

B.) 12 radians per sec^2

C.) 18 radians per sec^2

D.) 24 radians per sec^2 << my choice.

- Calculus PLEASE check my work , -
**MathMate**, Saturday, March 12, 2011 at 8:58am1. Correct

2. Correct

3.

dy/dx

=(dy/dt) / (dx/dt)

put

x=rcos(t)=3cos²(t)=

y=rsin(t)=3sin(t)cos(t)

Calculate dy/dx and evaluate at t=120° (2π/3) to get

dy/dx=-1/√3

=> θ=150°

Angle with radius vector=150°-120°=30°.

4.

There may have been a typo in the question. If the question is asking for θ=**2^t**-6t, the answers should not be in round numbers.

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